A transformer with a primary rated at 240 volts and a secondary rated at 120 volts is supplying a load pulling 10 amps. How many amps flow in the primary circuit?

A transformer with a primary rated at 240 volts and a secondary rated at 120 volts is supplying a load pulling 10 amps. How many amps flow in the primary circuit?

• A. 10 Amps
• B. 2 Amps
• C. 5 Amps
• D. 20 Amps

Answer: (C)

In analyzing the transformer scenario given, we can apply the principle of conservation of power, which states that the power in the primary side of the transformer should equal the power in the secondary side, assuming the transformer is operating at ideal conditions with no losses. The power on the secondary side is calculated using the formula: \[ P_{secondary} = V_{secondary} \times I_{secondary} \] Substituting in the provided values: \[ P_{secondary} = 120 \text{ volts} \times 10 \text{ amps} = 1200 \text{ watts} \] For the primary side, we know the voltage is 240 volts. To find the current flowing in the primary circuit, we rearrange the power formula: \[ P_{primary} = V_{primary} \times I_{primary} \] Setting the primary power equal to the secondary power gives us: \[ 1200 \text{ watts} = 240 \text{ volts} \times I_{primary} \] Now we solve for \( I_{primary} \): \[ I_{primary} = \frac{1200 \text{ watts}}{240 \text{ volts}} = 5 \text{ amps

In analyzing the transformer scenario given, we can apply the principle of conservation of power, which states that the power in the primary side of the transformer should equal the power in the secondary side, assuming the transformer is operating at ideal conditions with no losses.

The power on the secondary side is calculated using the formula:

[ P_{secondary} = V_{secondary} \times I_{secondary} ]

Substituting in the provided values:

[ P_{secondary} = 120 \text{ volts} \times 10 \text{ amps} = 1200 \text{ watts} ]

For the primary side, we know the voltage is 240 volts. To find the current flowing in the primary circuit, we rearrange the power formula:

[ P_{primary} = V_{primary} \times I_{primary} ]

Setting the primary power equal to the secondary power gives us:

[ 1200 \text{ watts} = 240 \text{ volts} \times I_{primary} ]

Now we solve for ( I_{primary} ):

[ I_{primary} = \frac{1200 \text{ watts}}{240 \text{ volts}} = 5 \text{ amps